# NCERT Solutions for Exercise 4.6 Class 12 Maths Chapter 4 - Determinants

In this article, you will get NCERT solutions for Class 12 Maths chapter 4 exercise 4.6 consists of questions related to applications of matrix and determinants. There are a lot of applications of matrices and determinants in the field of research, statistics, computer graphics, business, economics, and mathematics. In exercise 4.6 Class 12 Maths you will learn the specific application of matrix and determinant i.e. solving system of linear equation using the inverse of the matrix. All the questions in Class 12 Maths ch 4 ex 4.6 are related to solving the system of linear equations having unique solutions. In this Class 12 Maths chapter 4 exercise 4.6 solutions you will get questions related to finding consistency of linear equations. Also, check NCERT Solutions here.

Also, see

• Determinants Exercise 4.1
• Determinants Exercise 4.2
• Determinants Exercise 4.3
• Determinants Exercise 4.4
• Determinants Exercise 4.5
• Determinants Miscellaneous Exercise

Determinants Exercise:4.6

Question:1 Examine the consistency of the system of equations.

$\small x+2y=2$

$\small 2x+3y=3$

We have given the system of equations:18967

$\small x+2y=2$

$\small 2x+3y=3$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A= \begin{bmatrix} 1 &2 \\ 2&3 \end{bmatrix}$, $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 2\\3 \end{bmatrix}$.

So, we want to check for the consistency of the equations;

$|A| = 1(3) -2(2) = -1 \neq 0$

Here A is non -singular therefore there exists $A^{-1}$.

Hence, the given system of equations is consistent.

Question:2 Examine the consistency of the system of equations

$\small 2x-y=5$

$\small x+y=4$

We have given the system of equations:

$\small 2x-y=5$

$\small x+y=4$

The given system of equations can be written in the form of matrix; $AX =B$

where $A= \begin{bmatrix} 2 &-1 \\ 1&1 \end{bmatrix}$, $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 5\\4 \end{bmatrix}$.

So, we want to check for the consistency of the equations;

$|A| = 2(1) -1(-1) = 3 \neq 0$

Here A is non -singular therefore there exists $A^{-1}$.

Hence, the given system of equations is consistent.

Question:3 Examine the consistency of the system of equations.

$\small x+3y=5$

$\small 2x+6y=8$

We have given the system of equations:

$\small x+3y=5$

$\small 2x+6y=8$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A= \begin{bmatrix} 1 &3 \\ 2&6 \end{bmatrix}$, $X= \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 5\\8 \end{bmatrix}$.

So, we want to check for the consistency of the equations;

$|A| = 1(6) -2(3) = 0$

Here A is singular matrix therefore now we will check whether the $(adjA)B$ is zero or non-zero.

$adjA= \begin{bmatrix} 6 &-3 \\ -2& 1 \end{bmatrix}$

So, $(adjA)B= \begin{bmatrix} 6 &-3 \\ -2& 1 \end{bmatrix}\begin{bmatrix} 5\\8 \end{bmatrix} = \begin{bmatrix} 30-24\\-10+8 \end{bmatrix}=\begin{bmatrix} 6\\-2 \end{bmatrix} \neq 0$

As, $(adjA)B \neq 0$ , the solution of the given system of equations does not exist.

Hence, the given system of equations is inconsistent.

Question:4 Examine the consistency of the system of equations.

$\small x+y+z=1$

$\small 2x+3y+2z=2$

$\small ax+ay+2az=4$

We have given the system of equations:

$\small x+y+z=1$

$\small 2x+3y+2z=2$

$\small ax+ay+2az=4$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A = \begin{bmatrix} 1& 1&1 \\ 2& 3& 2\\ a& a &2a \end{bmatrix}$, $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 1\\2 \\ 4 \end{bmatrix}$.

So, we want to check for the consistency of the equations;

$|A| = 1(6a-2a) -1(4a-2a)+1(2a-3a)$

$= 4a -2a-a = 4a -3a =a \neq 0$

[If zero then it won't satisfy the third equation]

Here A is non- singular matrix therefore there exist $A^{-1}$.

Hence, the given system of equations is consistent.

Question:5 Examine the consistency of the system of equations.

$\small 3x-y-2z=2$

$\small 2y-z=-1$

$\small 3x-5y=3$

We have given the system of equations:

$\small 3x-y-2z=2$

$\small 2y-z=-1$

$\small 3x-5y=3$

The given system of equations can be written in the form of matrix; $AX =B$

where $A = \begin{bmatrix} 3& -1&-2 \\ 0& 2& -1\\ 3& -5 &0 \end{bmatrix}$, $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 2\\-1 \\ 3 \end{bmatrix}$.

So, we want to check for the consistency of the equations;

$|A| = 3(0-5) -(-1)(0+3)-2(0-6)$

$= -15 +3+12 = 0$

Therefore matrix A is a singular matrix.

So, we will then check $(adjA)B,$

$(adjA) = \begin{bmatrix} -5 &10 &5 \\ -3& 6 & 3\\ -6& 12 & 6 \end{bmatrix}$

$\therefore (adjA)B = \begin{bmatrix} -5 &10 &5 \\ -3& 6 & 3\\ -6& 12 & 6 \end{bmatrix}\begin{bmatrix} 2\\-1 \\ 3 \end{bmatrix} = \begin{bmatrix} -10-10+15\\ -6-6+9 \\ -12-12+18 \end{bmatrix} = \begin{bmatrix} -5\\-3 \\ -6 \end{bmatrix} \neq 0$

As, $(adjA)B$ is non-zero thus the solution of the given system of the equation does not exist. Hence, the given system of equations is inconsistent.

Question:6 Examine the consistency of the system of equations.

$\small 5x-y+4z=5$

$\small 2x+3y+5z=2$

$\small 5x-2y+6z=-1$

We have given the system of equations:

$\small 5x-y+4z=5$

$\small 2x+3y+5z=2$

$\small 5x-2y+6z=-1$

The given system of equations can be written in the form of the matrix; $AX =B$

where $A = \begin{bmatrix} 5& -1&4 \\ 2& 3& 5\\ 5& -2 &6 \end{bmatrix}$, $X = \begin{bmatrix} x\\y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 5\\2 \\ -1 \end{bmatrix}$.

So, we want to check for the consistency of the equations;

$|A| = 5(18+10) +1(12-25)+4(-4-15)$

$= 140-13-76 = 51 \neq 0$

Here A is non- singular matrix therefore there exist $A^{-1}$.

Hence, the given system of equations is consistent.

Question:7 Solve system of linear equations, using matrix method.

$\small 5x+2y=4$

$\small 7x+3y=5$

The given system of equations

$\small 5x+2y=4$

$\small 7x+3y=5$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 5 &4 \\ 7& 3 \end{bmatrix}$, $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 4\\5 \end{bmatrix}$

we have,

$|A| = 15-14=1 \neq 0$.

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = (adjA) = \begin{bmatrix} 3 &-2 \\ -7& 5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \begin{bmatrix} 3 &-2 \\ -7 & 5 \end{bmatrix}\begin{bmatrix} 4\\5 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} 12-10\\ -28+25 \end{bmatrix} = \begin{bmatrix} 2\\-3 \end{bmatrix}$

Hence the solutions of the given system of equations;

x = 2 and y =-3.

Question:8 Solve system of linear equations, using matrix method.

$2x-y=-2$

$3x+4y=3$

The given system of equations

$2x-y=-2$

$3x+4y=3$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &-1 \\ 3& 4 \end{bmatrix}$, $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} -2\\3 \end{bmatrix}$

we have,

$|A| = 8+3=11 \neq 0$.

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{11}\begin{bmatrix} 4 &1 \\ -3& 2 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{11}\begin{bmatrix} 4 &1 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} -2\\3 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -8+3\\ 6+6 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -5\\12 \end{bmatrix}= \begin{bmatrix} -\frac{5}{11}\\ \\-\frac{12}{11} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =\frac{-5}{11} \ and\ y =\frac{12}{11}.$

Question:9 Solve system of linear equations, using matrix method.

$\small 4x-3y=3$

$\small 3x-5y=7$

The given system of equations

$\small 4x-3y=3$

$\small 3x-5y=7$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 4 &-3 \\ 3& -5 \end{bmatrix}$, $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 3\\7 \end{bmatrix}$

we have,

$|A| = -20+9=-11 \neq 0$.

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{-1}{11}\begin{bmatrix} -5 &3 \\ -3& 4 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 5 &-3 \\ 3& -4 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{11}\begin{bmatrix} 5 &-3 \\ 3 & -4 \end{bmatrix}\begin{bmatrix} 3\\7 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{11}\begin{bmatrix} 15-21\\ 9-28 \end{bmatrix} = \frac{1}{11}\begin{bmatrix} -6\\-19 \end{bmatrix}= \begin{bmatrix} -\frac{6}{11}\\ \\-\frac{19}{11} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =\frac{-6}{11} \ and\ y =\frac{-19}{11}.$

Question:10 Solve system of linear equations, using matrix method.

$\small 5x+2y=3$

$\small 3x+2y=5$

The given system of equations

$\small 5x+2y=3$

$\small 3x+2y=5$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 5 &2 \\ 3& 2 \end{bmatrix}$, $X = \begin{bmatrix} x\\y \end{bmatrix}$ and $B = \begin{bmatrix} 3\\5 \end{bmatrix}$

we have,

$|A| = 10-6=4 \neq 0$.

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{4}\begin{bmatrix} 2 &-2 \\ -3& 5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{4}\begin{bmatrix} 2 &-2 \\ -3 & 5 \end{bmatrix}\begin{bmatrix} 3\\5 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x\\y \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 6-10\\ -9+25 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} -4\\16 \end{bmatrix}= \begin{bmatrix} -1\\4 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =-1 \ and\ y =4.$

Question:11 Solve system of linear equations, using matrix method.

$\small 2x+y+z=1$

$\small x-2y-z= \frac{3}{2}$

$\small 3y-5z=9$

The given system of equations

$\small 2x+y+z=1$

$\small x-2y-z= \frac{3}{2}$

$\small 3y-5z=9$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &1 &1 \\ 1 & -2 &-1 \\ 0& 3 &-5 \end{bmatrix}$, $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ and $B =\begin{bmatrix} 1\\ \\ \frac{3}{2} \\ \\ 9 \end{bmatrix}$

we have,

$|A| =2(10+3)-1(-5-0)+1(3-0) = 26+5+3 = 34 \neq 0$.

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(10+3) = 13$ $A_{12} =(-1)^{1+2}(-5-0) = 5$

$A_{13} =(-1)^{1+3}(3-0) = 3$ $A_{21} =(-1)^{2+1}(-5-3) = 8$

$A_{22} =(-1)^{2+2}(-10-0) = -10$ $A_{23} =(-1)^{2+3}(6-0) = -6$

$A_{31} =(-1)^{3+1}(-1+2) = 1$ $A_{32} =(-1)^{3+2}(-2-1) = 3$

$A_{33} =(-1)^{3+3}(-4-1) = -5$

$(adjA) =\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{34}\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \frac{1}{34}\begin{bmatrix} 13 &8 &1 \\ 5& -10 & 3\\ 3& -6 & -5 \end{bmatrix}\begin{bmatrix} 1\\\frac{3}{2} \\ 9 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{34}\begin{bmatrix} 13+12+9\\5-15+27 \\ 3-9-45 \end{bmatrix} = \frac{1}{34}\begin{bmatrix} 34\\17 \\ -51 \end{bmatrix}= \begin{bmatrix} 1\\\frac{1}{2} \\ -\frac{3}{2} \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =\frac{1}{2},\ and\ \ z=-\frac{3}{2}.$

Question:12 Solve system of linear equations, using matrix method.

$\small x-y+z=4$

$\small 2x+y-3z=0$

$\small x+y+z=2$

The given system of equations

$\small x-y+z=4$

$\small 2x+y-3z=0$

$\small x+y+z=2$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 1 &-1 &1 \\ 2 & 1 &-3 \\ 1& 1 &1 \end{bmatrix}$, $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 4\\ 0 \\ 2 \end{bmatrix}.$

we have,

$|A| =1(1+3)+1(2+3)+1(2-1) = 4+5+1= 10 \neq 0$.

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(1+3) = 4$ $A_{12} =(-1)^{1+2}(2+3) = -5$

$A_{13} =(-1)^{1+3}(2-1) = 1$ $A_{21} =(-1)^{2+1}(-1-1) = 2$

$A_{22} =(-1)^{2+2}(1-1) = 0$ $A_{23} =(-1)^{2+3}(1+1) = -2$

$A_{31} =(-1)^{3+1}(3-1) = 2$ $A_{32} =(-1)^{3+2}(-3-2) = 5$

$A_{33} =(-1)^{3+3}(1+2) = 3$

$(adjA) =\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{10}\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{10}\begin{bmatrix} 4 &2 &2 \\ -5& 0 & 5\\ 1& -2 & 3 \end{bmatrix}\begin{bmatrix} 4\\0 \\ 2 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 16+0+4\\-20+0+10 \\ 4+0+6 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 20\\-10 \\ 10 \end{bmatrix}= \begin{bmatrix} 2\\-1 \\ 1 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =2,\ y =-1,\ and\ \ z=1.$

Question:13 Solve system of linear equations, using matrix method.

$\small 2x+3y+3z=5$

$\small x-2y+z=-4$

$\small 3x-y-2z=3$

The given system of equations

$\small 2x+3y+3z=5$

$\small x-2y+z=-4$

$\small 3x-y-2z=3$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 2 &3 &3 \\ 1 & -2 &1 \\ 3& -1 &-2 \end{bmatrix}$, $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 5\\ -4 \\ 3 \end{bmatrix}.$

we have,

$|A| =2(4+1) -3(-2-3)+3(-1+6) = 10+15+15 = 40$.

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{1+1}(4+1) = 5$ $A_{12} =(-1)^{1+2}(-2-3) = 5$

$A_{13} =(-1)^{1+3}(-1+6) = 5$ $A_{21} =(-1)^{2+1}(-6+3) = 3$

$A_{22} =(-1)^{2+2}(-4-9) = -13$ $A_{23} =(-1)^{2+3}(-2-9) = 11$

$A_{31} =(-1)^{3+1}(3+6) = 9$ $A_{32} =(-1)^{3+2}(2-3) = 1$

$A_{33} =(-1)^{3+3}(-4-3) = -7$

$(adjA) =\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5&11 & -7 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{40}\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5& 11 & -7 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{40}\begin{bmatrix} 5 &3 &9 \\ 5& -13 & 1\\ 5& 11 & -7 \end{bmatrix}\begin{bmatrix} 5\\-4 \\ 3 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 25-12+27\\25+52+3 \\ 25-44-21 \end{bmatrix} = \frac{1}{40}\begin{bmatrix} 40\\80 \\ -40 \end{bmatrix}= \begin{bmatrix} 1\\2 \\ -1 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =2,\ and\ \ z=-1.$

Question:14 Solve system of linear equations, using matrix method.

$\small x-y+2z=7$

$\small 3x+4y-5z=-5$

$\small 2x-y+3z=12$

The given system of equations

$\small x-y+2z=7$

$\small 3x+4y-5z=-5$

$\small 2x-y+3z=12$

can be written in the matrix form of AX =B, where

$A = \begin{bmatrix} 1 &-1 &2 \\ 3 & 4 &-5 \\ 2& -1 &3 \end{bmatrix}$, $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 7\\ -5 \\ 12 \end{bmatrix}.$

we have,

$|A| =1(12-5) +1(9+10)+2(-3-8) = 7+19-22 = 4 \neq0$.

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{12-5} = 7$ $A_{12} =(-1)^{1+2}(9+10) = -19$

$A_{13} =(-1)^{1+3}(-3-8) = -11$ $A_{21} =(-1)^{2+1}(-3+2) = 1$

$A_{22} =(-1)^{2+2}(3-4) = -1$ $A_{23} =(-1)^{2+3}(-1+2) = -1$

$A_{31} =(-1)^{3+1}(5-8) = -3$ $A_{32} =(-1)^{3+2}(-5-6) = 11$

$A_{33} =(-1)^{3+3}(4+3) = 7$

$(adjA) =\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{4}\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B =\frac{1}{4}\begin{bmatrix} 7 &1 &-3 \\ -19& -1 & 11\\ -11&-1 & 7 \end{bmatrix}\begin{bmatrix} 7\\-5 \\ 12 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 49-5-36\\-133+5+132 \\ -77+5+84 \end{bmatrix} = \frac{1}{4}\begin{bmatrix} 8\\4 \\ 12 \end{bmatrix}= \begin{bmatrix} 2\\1 \\ 3 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =2,\ y =1,\ and\ \ z=3.$

Question:15 If $\dpi{100} A=\begin{bmatrix} 2 &-3 &5 \\ 3 & 2 &-4 \\ 1 &1 &-2 \end{bmatrix}$ , find $\dpi{100} A^-^1$. Using $\dpi{100} A^-^1$ solve the system of equations

$\dpi{100} 2x-3y+5z=11$

$\dpi{100} 3x+2y-4z=-5$

$\dpi{100} x+y-2z=-3$

The given system of equations

$\dpi{100} 2x-3y+5z=11$

$\dpi{100} 3x+2y-4z=-5$

$\dpi{100} x+y-2z=-3$

can be written in the matrix form of AX =B, where

$\dpi{100} A=\begin{bmatrix} 2 &-3 &5 \\ 3 & 2 &-4 \\ 1 &1 &-2 \end{bmatrix}$, $X = \begin{bmatrix} x\\y \\z \end{bmatrix}$ $and\ B =\begin{bmatrix} 11\\ -5 \\ -3 \end{bmatrix}.$

we have,

$|A| =2(-4+4) +3(-6+4)+5(3-2) = 0-6+5 = -1 \neq0$.

So, A is non-singular, Therefore, its inverse $A^{-1}$ exists.

as we know $A^{-1} = \frac{1}{|A|} (adjA)$

Now, we will find the cofactors;

$A_{11} =(-1)^{-4+4} = 0$ $A_{12} =(-1)^{1+2}(-6+4) = 2$

$A_{13} =(-1)^{1+3}(3-2) = 1$ $A_{21} =(-1)^{2+1}(6-5) = -1$

$A_{22} =(-1)^{2+2}(-4-5) = -9$ $A_{23} =(-1)^{2+3}(2+3) = -5$

$A_{31} =(-1)^{3+1}(12-10) = 2$ $A_{32} =(-1)^{3+2}(-8-15) = 23$

$A_{33} =(-1)^{3+3}(4+9) = 13$

$(adjA) =\begin{bmatrix} 0 &-1 &2 \\ 2& -9 & 23\\ 1&-5 & 13 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = -1\begin{bmatrix} 0 &-1 &2 \\ 2& -9 & 23\\ 1&-5 & 13 \end{bmatrix} = \begin{bmatrix} 0 &1 &-2 \\ -2& 9 & -23\\ -1&5 & -13 \end{bmatrix}$

So, the solutions can be found by $X = A^{-1}B = \begin{bmatrix} 0 &1 &-2 \\ -2& 9 & -23\\ -1&5 & -13 \end{bmatrix}\begin{bmatrix} 11\\-5 \\ -3 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 0-5+6\\-22-45+69 \\ -11-25+39 \end{bmatrix} = \begin{bmatrix} 1\\2 \\ 3 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =1,\ y =2,\ and\ \ z=3.$

Question:16 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg by matrix method.

So, let us assume the cost of onion, wheat, and rice be x, y and z respectively.

Then we have the equations for the given situation :

$4x+3y+2z = 60$

$2x+4y+6z = 90$

$6x+2y+3y = 70$

We can find the cost of each item per Kg by the matrix method as follows;

Taking the coefficients of x, y, and z as a matrix $A$.

We have;

$A = \begin{bmatrix} 4 &3 &2 \\ 2& 4 &6 \\ 6 & 2 & 3 \end{bmatrix},$ $X= \begin{bmatrix} x\\y \\ z \end{bmatrix}$ $and\ B = \begin{bmatrix} 60\\90 \\ 70 \end{bmatrix}.$

$|A| = 4(12-12) -3(6-36)+2(4-24) = 0 +90-40 = 50 \neq 0$

Now, we will find the cofactors of A;

$A_{11} = (-1)^{1+1}(12-12) = 0$ $A_{12} = (-1)^{1+2}(6-36) = 30$

$A_{13} = (-1)^{1+3}(4-24) = -20$ $A_{21} = (-1)^{2+1}(9-4) = -5$

$A_{22} = (-1)^{2+2}(12-12) = 0$ $A_{23} = (-1)^{2+3}(8-18) = 10$

$A_{31} = (-1)^{3+1}(18-8) = 10$ $A_{32} = (-1)^{3+2}(24-4) = -20$

$A_{33} = (-1)^{3+3}(16-6) = 10$

$adjA = \begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}$

$A^{-1} = \frac{1}{|A|} (adjA) = \frac{1}{50}\begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}$s

So, the solutions can be found by $X = A^{-1}B = \frac{1}{50}\begin{bmatrix} 0 &-5 &10 \\ 30 & 0 &-20 \\ -20 & 10 & 10 \end{bmatrix}\begin{bmatrix} 60\\90 \\ 70 \end{bmatrix}$

$\Rightarrow\begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 0-450+700\\1800+0-1400 \\ -1200+900+700 \end{bmatrix} =\frac{1}{50} \begin{bmatrix} 250\\400 \\ 400 \end{bmatrix} = \begin{bmatrix} 5\\8 \\ 8 \end{bmatrix}$

Hence the solutions of the given system of equations;

$x =5,\ y =8,\ and\ \ z=8.$

Therefore, we have the cost of onions is Rs. 5 per Kg, the cost of wheat is Rs. 8 per Kg, and the cost of rice is Rs. 8 per kg.

## More About NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6:-

In Class 12th Maths chapter 4 exercise 4.6 solutions, there are 16 long answer-type questions which you should try to solve by yourself. There are three solved examples given in the NCERT textbook before exercise 4.6 Class 12 Maths that you can solve. Solving these examples will help you in solving this exercise 4.6 Class 12 Maths. This chapter is important for the board exams as well as for the higher studies in engineering. Solving a system of linear equations is a must-be-know tool for any engineering branch. You are advised to go through these NCERT Solutions for Class 12 Maths chapter 4 Exercise 4.6 for conceptual clarity.

Also Read| Determinants Class 12 Chapter 4 Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6:-

• This chapter is very important for board exams and competitive exams like JEE, SRMJEE, etc.
• Class 12 Maths chapter 4 exercise 4.6 solutions are helpful for you in solving the system of linear equations and linear equations consistency problems.
• NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 are helpful for the students who are not able to solve NCERT problems by themself.
• In the Class 12th Maths chapter 4 exercise 4.6 solutions, you will learn the art of answering in board exams to get good marks.
• You can take these Class 12 Maths chapter 4 exercise 4.6 solutions for reference.

## Also see-

• NCERT Solutions for Class 12 Maths Chapter 4

• NCERT Exemplar Solutions Class 12 Maths Chapter 4

## NCERT Solutions of Class 12 Subject Wise

• NCERT Solutions for Class 12 Maths
• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Chemistry

• NCERT Solutions for Class 12 Biology

## Subject Wise NCERT Exampler Solutions

• NCERT Exemplar Solutions for Class 12th Maths

• NCERT Exemplar Solutions for Class 12th Physics

• NCERT Exemplar Solutions for Class 12th Chemistry

• NCERT Exemplar Solutions for Class 12th Biology

Happy learning!!!